Correct Answer - Option 2 : 0
Concept:
Scalar Triple Product:
If \(\vec a = {a_1}\hat i + {a_2}\hat j + {a_3}\hat k\), \(\vec b = {b_1}\hat i + {b_2}\hat j + {b_3}\hat k\) and \(\vec c = {c_1}\hat i + {c_2}\hat j + {c_3}\hat k\) then their scalar triple product is defined as \(\vec a \cdot \left( {\vec b \times \;\vec c} \right) = \left| {\begin{array}{*{20}{c}} {{a_1}}&{{a_2}}&{{a_3}}\\ {{b_1}}&{{b_2}}&{{b_3}}\\ {{c_1}}&{{c_2}}&{{c_3}} \end{array}} \right| = \left[ {a\;b\;c} \right]\)
Calculation:
Given: \(\vec a = \hat i + \hat j + \hat k\;,\;\vec b = 6\hat i + 6\hat j + 6\hat k\;and\;\vec c = 2\hat i + 3\hat j + \hat k\)
As we know that, the scalar triple product of three vectors is \(\vec a \cdot \left( {\vec b \times \;\vec c} \right) = \left| {\begin{array}{*{20}{c}} {{a_1}}&{{a_2}}&{{a_3}}\\ {{b_1}}&{{b_2}}&{{b_3}}\\ {{c_1}}&{{c_2}}&{{c_3}} \end{array}} \right| = \left[ {a\;b\;c} \right]\)
⇒ \(\vec a \cdot \left( {\vec b \times \;\vec c} \right) = \left| {\begin{array}{*{20}{c}} {{1}}&{{1}}&{{1}}\\ {{6}}&{{6}}&{{6}}\\ {{2}}&{{3}}&{{1}} \end{array}} \right|\)
⇒ \(\vec a \cdot \left( {\vec b \times \;\vec c} \right) = 1 \times (6 - 18) - 1 \times (6 - 12) + 1 \times (18 - 12) = 0\)
Hence, the correct option is 2.
